Kapalci

Answered

2022-06-25

The form of the partial fraction decomposition of a rational function is given below.

$\frac{x-3{x}^{2}-26}{(x+1)({x}^{2}+9)}=\frac{A}{x+1}+\frac{Bx+C}{{x}^{2}+9}$

What are the values of $A,B$ and $C$?

So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$

I also found the indefinite integral, which equals

$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)-3\mathrm{log}(1+x)+C$

$\frac{x-3{x}^{2}-26}{(x+1)({x}^{2}+9)}=\frac{A}{x+1}+\frac{Bx+C}{{x}^{2}+9}$

What are the values of $A,B$ and $C$?

So can someone explain how to get the answer. I'm not sure how I got to the answer but I know $B=0$ and $C=1$

I also found the indefinite integral, which equals

$\frac{1}{3}{\mathrm{tan}}^{-1}\left(\frac{x}{3}\right)-3\mathrm{log}(1+x)+C$

Answer & Explanation

Lamont Adkins

Expert

2022-06-26Added 11 answers

Multiplying either sides by $(x+1)({x}^{2}+9),$

$x-3{x}^{2}-26=A({x}^{2}+9)+(Bx+C)(x+1)$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-3{x}^{2}+x-26={x}^{2}(A+B)+x(B+C)+9A+C$

Compare the constants and the coefficients of $x,{x}^{2}$ of the above identity to form three simultaneous equations with three unknowns A,B,C

$x-3{x}^{2}-26=A({x}^{2}+9)+(Bx+C)(x+1)$

$\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}-3{x}^{2}+x-26={x}^{2}(A+B)+x(B+C)+9A+C$

Compare the constants and the coefficients of $x,{x}^{2}$ of the above identity to form three simultaneous equations with three unknowns A,B,C

George Bray

Expert

2022-06-27Added 12 answers

Multiply by $x+1$ and let $x\to -1$ to obtain

$\frac{-1-3-26}{(-1{)}^{2}+9}=A\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=-3.$

Multiply both sides by $x$, let $x\to \mathrm{\infty}$, and use $A=-3$ to obtain

$-3=A+B\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B=0.$

Finally, evaluate at $x=0$, and use $A=-27/9$:

$\frac{-26}{9}=A+\frac{C}{9}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=9(-\frac{26}{9}+\frac{27}{9})=1.$

$\frac{-1-3-26}{(-1{)}^{2}+9}=A\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}A=-3.$

Multiply both sides by $x$, let $x\to \mathrm{\infty}$, and use $A=-3$ to obtain

$-3=A+B\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}B=0.$

Finally, evaluate at $x=0$, and use $A=-27/9$:

$\frac{-26}{9}=A+\frac{C}{9}\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}C=9(-\frac{26}{9}+\frac{27}{9})=1.$

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